A) \[AlC{{l}_{3}}>{{K}_{3}}\left[ Fe{{\left( CN \right)}_{6}} \right]>{{K}_{2}}Cr{{O}_{4}}>KBr=KN{{O}_{3}}\]
B) \[{{K}_{3}}\left[ Fe{{\left( CN \right)}_{6}} \right]<{{K}_{2}}Cr{{O}_{4}}<KBr=KN{{O}_{3}}=AlC{{l}_{3}}\]
C) \[{{K}_{3}}\left[ Fe{{\left( CN \right)}_{6}} \right]>AlC{{l}_{3}}>{{K}_{2}}Cr{{O}_{4}}>KBr>KN{{O}_{3}}\]
D) \[{{K}_{3}}\left[ Fe{{\left( CN \right)}_{6}} \right]<{{K}_{2}}Cr{{O}_{4}}<AlC{{l}_{3}}<KBr<KN{{O}_{3}}\]
Correct Answer: B
Solution :
[b] \[Fe{{\left( OH \right)}_{3}}\]is a positive sol. Its coagulation will be caused by the anion of the electrolyte. The flocculation value is inversely proportional to coagulation power or valence of the anion. The correct order of flocculation value is \[{{K}_{3}}\left[ Fe{{\left( CN \right)}_{6}} \right]<{{K}_{2}}Cr{{O}_{4}}<KBr=KN{{O}_{3}}=AlC{{l}_{3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
